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Torque at 6% slip from maximum torque

Short answer

Ns=1000 rpm; sm=(1000-960)/1000=0.04. Using T/Tmax=2 s sm/(s²+sm²) with s=0.06 gives T/Tmax=0.9231, so T=0.9231×30=27.69 N·m.

A three phase 6 pole, 50 Hz induction motor develops a maximum torque of 30 Nm at 960 rpm. Calculate the torque produced by the motor at 6% slip. The rotor resistance per phase is 0.6 Ω0.6\ \Omega.

Given: 6-pole, 50 Hz, Tmax=30T_{max} = 30 N-m at 960 rpm, R2=0.6 ΩR_2 = 0.6\ \Omega, s=0.06s = 0.06.

Ns=120×506=1000 rpmN_s = \frac{120\times 50}{6} = 1000\ \text{rpm}

Slip at maximum torque:

sm=1000−9601000=0.04s_m = \frac{1000-960}{1000} = 0.04

(This gives X2=R2/sm=15 ΩX_2 = R_2/s_m = 15\ \Omega at standstill, but the ratio method does not need it.)

Torque at any slip relative to maximum torque (neglecting stator impedance):

TTmax=2 s sms2+sm2\frac{T}{T_{max}} = \frac{2\,s\,s_m}{s^2 + s_m^2} TTmax=2×0.06×0.040.062+0.042=0.00480.0052=0.9231\frac{T}{T_{max}} = \frac{2\times 0.06\times 0.04}{0.06^2 + 0.04^2} = \frac{0.0048}{0.0052} = 0.9231 T=0.9231×30=27.69 N-mT = 0.9231\times 30 = 27.69\ \text{N-m}

Answer: torque at 6 % slip =27.69= 27.69 N-m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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